# Physics nerds: External ballistics quandry. **URL:** \ **Category:** Factual Questions\ **Created:** [January 28, 2017, 4:58am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368 "2017-01-28T04:58:00Z")\ **Posts on this page:** 11\ **Page:** 2
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 1:11am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/21 "2017-01-29T01:11:32Z")
> [@Riemann](#): > > So, if the amount of force acting to deflect the bullet from it’s aimed path is the full 1g of gravity when aiming horizonatlly, and zero when aiming vertically, when aiming at an intermediate diagonal angle uphill or downhill, it must be some intermediate figure less than 1g, right? So now see the diagram I posted. I **sincerely** appreciate your effort to explain. However, the idea of the actual force of gravity having 0% effect @ +90 and -90 degrees, and a proportional amount in between, doesn’t “click” with me as a non-physicist. The description given by\* Dr. Strangelove\* however, does. Perhaps because he? explains it in a manner of simple Geometry as related to the alignment of a scope in the horizontal, vs the same instrument utilized at an angle. Thank you for struggling with my ignorance. I appreciate it:) ---
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 1:17am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/22 "2017-01-29T01:17:32Z")
> [@Dr.Strangelove](#): > > But gravity works over time, not distance. And it still takes the same amount of time for the shot to get there, so it falls the same distance. > > It took me a few moments to figure out what’s going on. Ignoring velocity loss due to air resistance or gravity, the bullet will drop the same distance from its straight-line target regardless of the angle. But note that it _always drops straight down_. From the perspective of the scope, a straight-down drop from an angle appears smaller than it would be head-on. And conversely, a given amount of drop from the scope would correspond to a greater amount of physical drop if it happened to be at an angle. > > To be more concrete: suppose the rifle is 1500 f/s and the target is 300 feet away. The bullet will drop 0.64 feet at the end of its path regardless of angle. > > If looked at head-on, this makes an angle of 0.12 degrees, so you would point the scope down at that angle if calibrating for that muzzle velocity and distance. But if you were shooting up at 45 degrees, it would look like only 0.09 degrees. So the pre-compensation in the scope would be too much, and the shooter would have to point down a bit to correct for it. Thank you for explaining this from a perspective that a dumb-ass such as I, can understand. You have cleared a mental fog that I have had off and on, for many years!🙂 ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [January 29, 2017, 1:44am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/23 "2017-01-29T01:44:12Z")
> [@ZurBob](#): > > You have cleared a mental fog that I have had off and on, for many years!🙂 Glad to help! Don’t sell yourself short, though. There have been lots of times when I just didn’t get something until I hit on the right interpretation/explanation. Fortunately, in physics there’s usually a bunch of different, but equivalent ways of looking at a situation. Eventually, one of them makes sense, and often you can translate that intuition to other explanations once you get a foothold. ---
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 1:53am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/24 "2017-01-29T01:53:48Z")
> [@ZurBob](#): > > Just a brain fart here, BUT: If I’m on the right track… It would require a greater energy input to cause an object to travel a horizontal path of distance x, than the same object @ 45 degrees over distance x ? As I understand it now, in response to my own question, the answer would be _ **no.** _ ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [January 29, 2017, 1:57am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/25 "2017-01-29T01:57:19Z")
> [@ZurBob](#): > > …The description given by\* Dr. Strangelove\* however, does… If you “get” that way of looking at it, you can actually extract the way Dr Strangelove described things from the exact same diagram that I posted. Think of the 1g vector as representing the amount of deflection of the bullet in each case (exaggerated). In the horizontal case, the bullet hits the target the entire 1g off-center. In the diagonal cases, the dotted 0.7g perpendicular component represents how large the deflection _appears_ to someone looking through the rifle sight along the path of the bullet, or equivalently how far off-center the bullet will hit the target. ---
**Author:** ![eschereal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eschereal/32/18939_2.png) [@eschereal](https://boards.straightdope.com/u/eschereal)\ **Post date:** [January 29, 2017, 2:53am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/26 "2017-01-29T02:53:37Z")
> [@ZurBob](#): > > As I understand it now, in response to my own question, the answer would be _ **no.** _ When you shoot level, the slug begins descending as soon as it leaves the barrel. When you shoot at an upward angle, the slug’s horizontal motion is less, but it begins its journey rising, before it the starts descending. Up to 45°, the reduction in horizontal speed is made up for by the increased flight time: if your sight is set for 100yds but your target is 150yds away, level, you raise the barrel a tad to get your shot there (watch the battle scene from _Braveheart_ to see how shooting up inceases range – the same basic physics apply to arrows). Angles higher than 45° result in longer flight times offset by much lower horizontal speed, so the overall travel distance decreases. ---
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 3:05am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/27 "2017-01-29T03:05:36Z")
> [@Riemann](#): > > If you “get” that way of looking at it, you can actually extract the way Dr Strangelove described things from the exact same diagram that I posted. Think of the 1g vector as representing the amount of deflection of the bullet in each case (exaggerated). In the horizontal case, the bullet hits the target the entire 1g off-center. In the diagonal cases, the dotted 0.7g perpendicular component represents how large the deflection _appears_ to someone looking through the rifle sight along the path of the bullet, or equivalently how far off-center the bullet will hit the target. Well, yes, the math may work out the same. But isn’t the fundamental explanation different… As in “Optical geometry” vs “Differences in gravitational force based upon angular deflection” ? To me, the first is understandable… Your explanation forces the numbers to agree with a preconceived conclusion, and is more difficult for me to grasp. Sorry for the blunt aspect of my reply, no disrespect intended. ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [January 29, 2017, 3:51am UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/28 "2017-01-29T03:51:11Z")
> [@ZurBob](#): > > Well, yes, the math may work out the same. But isn’t the fundamental explanation different… No, it’s equivalent. > [@ZurBob](#): > > … Your explanation forces the numbers to agree with a preconceived conclusion… And no, it doesn’t. It’s elementary high school trigonometry. Perhaps you might want to reconsider such a confrontational discursive style when by your own admission you are clueless about what’s going here? ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [January 29, 2017, 12:19pm UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/29 "2017-01-29T12:19:04Z")
Understanding happens at both the “I can do the process to get the answer” level and the “I get it” level. When I find folks “blind” to an explanation it’s usually a matter that somewhere upstream they’ve built in a faulty assumption that’s [del]derailing[/del] utterly blocking understanding. If we can tease that out we have hope of moving forward. Until then the unidentified block induces nothing but inchoate frustration in the student. Kind of like how the ancients struggled with how gravity inconsistently affected feathers vs. anvils because they had unwittingly built air resistance into their model of gravity because they’d never seen anything fall in a vacuum. Once they realized they were seeing the sum of two effects, a consistent gravity plus an inconsistent air resistance, suddenly the light bulb \*could \*come on. So, as a first step in rooting out the blockage … Does the OP understand 2D vectors at the “I get it” level? Does he recognize that any vector can be decomposed into two components set at whatever angle to each other and to up/down, left/right, etc.? That any single decomposition that sums correctly is inherently no better or worse or more or less correct than any other? Certainly some are more useful n others to describe a particular situation. ---
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 2:02pm UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/30 "2017-01-29T14:02:03Z")
> [@Riemann](#): > > …Perhaps you might want to reconsider such a confrontational discursive style when by your own admission you are clueless about what’s going here? 😕 😕 You, and every one else here, have been very helpful. **Again,** thank you very much. ---
**Author:** ![ZurBob](https://avatars.discourse-cdn.com/v4/letter/z/71e660/32.png) [@ZurBob](https://boards.straightdope.com/u/ZurBob)\ **Post date:** [January 29, 2017, 2:15pm UTC](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368/31 "2017-01-29T14:15:26Z")
> [@LSLGuy](#): > > Understanding happens at both the “I can do the process to get the answer” level and the “I get it” level… > > Does the OP understand 2D vectors at the “I get it” level? Does he recognize that any vector can be decomposed into two components set at whatever angle to each other and to up/down, left/right, etc.? That any single decomposition that sums correctly is inherently no better or worse or more or less correct than any other? Certainly some are more useful n others to describe a particular situation. My exposure to vector analysis is very limited, so I had to arrive at the “I get it” level from another angle (no pun intended). Sorry if some feathers got ruffled in the process. I appreciate everyone’s effort to bring the slow kid up to speed.🙂 [Previous page](https://boards.straightdope.com/t/physics-nerds-external-ballistics-quandry/778368.md?page=1)